Integrand size = 28, antiderivative size = 125 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {\left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
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Time = 0.50 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {4482, 2968, 3127, 3110, 3100, 2827, 3852, 8, 3855} \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {\left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {a b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d} \]
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Rule 8
Rule 2827
Rule 2968
Rule 3100
Rule 3110
Rule 3127
Rule 3852
Rule 3855
Rule 4482
Rubi steps \begin{align*} \text {integral}& = \int (b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan ^2(c+d x) \, dx \\ & = \int (b+a \cos (c+d x))^2 \left (1-\cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx \\ & = \frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{12} \int \left (-3 \left (2 a^2-b^2\right )+8 a b \cos (c+d x)+9 a^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (16 a b+3 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{3} (2 a b) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (4 a^2+b^2\right ) \int \sec (c+d x) \, dx \\ & = -\frac {\left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(2 a b) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = -\frac {\left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.62 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {-3 \left (4 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2-b^2\right ) \sec (c+d x)+6 b^2 \sec ^3(c+d x)+16 a b \tan ^2(c+d x)\right )}{24 d} \]
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Time = 1.97 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(138\) |
default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {2 a b \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin \left (d x +c \right )^{3}}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(138\) |
risch | \(-\frac {i \left (12 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+48 a b \,{\mathrm e}^{6 i \left (d x +c \right )}+12 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+21 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+48 a b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-21 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+16 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2} {\mathrm e}^{i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{i \left (d x +c \right )}+16 a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2}}{2 d}-\frac {b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{2 d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(261\) |
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Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {3 \, {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, a b \cos \left (d x + c\right )^{3} - 16 \, a b \cos \left (d x + c\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
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\[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]
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Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {32 \, a b \tan \left (d x + c\right )^{3} + 3 \, b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]
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Time = 1.00 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.81 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {3 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 64 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 64 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
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Time = 25.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.42 \[ \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {\left (a^2+\frac {b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-a^2-\frac {16\,a\,b}{3}+\frac {7\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-a^2+\frac {16\,a\,b}{3}+\frac {7\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^2+\frac {b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+\frac {b^2}{4}\right )}{d} \]
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